UMD Analysis Qualifying Exam/Aug12 Complex

Problem 4

Suppose $h(z)$ is holomorphic on a region containing the disk $\{|z|\le R\}$ and that $|h(z)|<R$ if $|z|=R$. How many solutions does the equation $h(z)=z$ have in the disc $\{|z|\le R\}$? Justify your answer.

Solution 4

We know $|h(z)|=|h(z)-z+z|<R$ on $|z|=R$. Similarly, since $|h(z)-z|\ge 0$ then $|h(z)-z|+|z|\ge R$ on $|z|=R$. This gives $|h(z)-z+z|<R\le |h(z)-z|+|z|$ on $z=R$.

So by Rouché's theorem, since both functions are holomorphic (i.e. have no poles), then $h(z)-z$ has the same number of zeros as $z$ on the domain $|z|<R$. Since $z$ has only one zero (namely 0), then there is only one solution to $h(z)=z$ inside the open disc $|z|<R$.

Observe that $h(z)\ne z$ for any $|z|=R$, since that would imply $|h(z)|=R$ for some $z$ on the boundary, contradicting the hypothesis.

Thus, there is only one solution to $h(z)=z$ inside the open disc $|z|\le R$.